Question: $ \left(\dfrac{1}{16}\right)^{-\frac{3}{2}}$
$= 16^{\frac{3}{2}}$ $= \left(16^{\frac{1}{2}}\right)^{3}$ To simplify $16^{\frac{1}{2}}$ , figure out what goes in the blank: $\left(? \right)^{2}=16$ To simplify $16^{\frac{1}{2}}$ , figure out what goes in the blank: $\left({4}\right)^{2}=16$ so $ 16^{\frac{1}{2}}=4$ So $16^{\frac{3}{2}}=\left(16^{\frac{1}{2}}\right)^{3}=4^{3}$ $= 4^{3}$ $= 4\cdot4\cdot 4$ $= 16\cdot4$ $= 64$